1964 AHSME Problems/Problem 29

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Problem

In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is:

$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$

[asy] import olympiad;  draw((0,0)--7.5*dir(0)); draw((0,0)--5*dir(55)); draw((0,0)--4*dir(140)); draw(4*dir(140)--5*dir(55)--7.5*dir(0)); markscalefactor=0.1; draw(anglemark((0,0),4*dir(140),5*dir(55))); draw(anglemark(7.5*dir(0),(0,0),5*dir(55))); label("$F$",(0,0),dir(240)); label("$D$",4*dir(140),dir(180)); label("$R$",5*dir(55),dir(80)); label("$S$",7.5*dir(0),dir(-25)); label("$4$",2*dir(140),dir(230)); label("$5$",2.5*dir(55),dir(155)); label("$6$",(4*dir(140)+5*dir(55))/2,dir(100)); label("$7\frac{1}{2}$",3.75,dir(-90)); [/asy]

Solution

We examine $\triangle RDF$ and $\triangle SFR$. We are given $\angle RDF \cong \angle SFR$. Also note that $\frac{SF}{RD} = \frac{7.5}{6} = 1.25$ and $\frac{FR}{DF} = \frac{5}{4} = 1.25$, so $\frac{SF}{RD} = \frac{FR}{DF}$.

If two triangles have two pairs of sides that are proportional, and the included angle is congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so

$\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25$

$\frac{RS}{FR} = 1.25$

$\frac{RS}{5} = 1.25$

$RS = 6.25$, which is answer $\boxed{\textbf{(E) }}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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